不光是查找值！"二分搜索"

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从有序数组中查找某个值

• 问题描述：给定长度为n的单调不下降数列a₀,…,a_n-1和一个数k，求满足a_i≥k条件的最小的i。不存在则输出n。
• 限制条件：
  1≤n≤10⁶
  0≤a₀≤a₁≤…≤a_n-1<10⁹
  0≤k≤10⁹
• 分析：二分搜索。STL以lower_bound函数的形式实现了二分搜索。
• 代码：

   #include <cstdio>
   #include <cctype>
   #include <algorithm>
   #define num s-'0'
  
   using namespace std;
  
   const int MAX_N=;
   const int INF=0x3f3f3f3f;
   int n,k;
   int a[MAX_N];
  
   void read(int &x){
       char s;
       x=;
       bool flag=;
       while(!isdigit(s=getchar()))
           (s=='-')&&(flag=true);
       for(x=num;isdigit(s=getchar());x=x*+num);
       (flag)&&(x=-x);
   }
  
   void write(int x)
   {
       if(x<)
       {
           putchar('-');
           x=-x;
       }
       if(x>)
           write(x/);
       putchar(x%+'');
   }
  
   int search(int);
  
   int main()
   {
       read(n);read(k);
       for (int i=; i<n; i++) read(a[i]);
       int p = search(k);
       write(p);
       putchar('\n');
       write(lower_bound(a,a+n,k)-a);
       putchar('\n');
   }
  
   int search(int k)
   {
       int l=-, r=n-;
       while (r-l>)
       {
           int mid=(r+l)/;
           if (a[mid]>=k) r=mid;
           else l=mid;
       }
       return r;
   }

  lower_bound

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假定一个解并判断是否可行

对于任意满足C(x)的x，如果所有的x'≥x也满足C(x')的话，我们就可以用二分搜索来求得最小的x。首先我们将区间的左端点初始化为不满足C(x)的值，右端点初始化为满足C(x)的值，然后每次取中点mid=(lb+ub)/2，判断C(mid)是否满足并缩小范围，直到(lb,ub]足够小了为止，最后ub就是要求的最小值。最大化的问题也可以用同样的方法求解。

Cable master(POJ 1064)

• 原题如下：
  Cable master

  Time Limit: 1000MS		Memory Limit: 10000K
  Total Submissions: 65114		Accepted: 13413

  Description

  Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to organize the most honest contest ever. It was decided to connect computers for the contestants using a "star" topology - i.e. connect them all to a single central hub. To organize a truly honest contest, the Head of the Judging Committee has decreed to place all contestants evenly around the hub on an equal distance from it. 
  To buy network cables, the Judging Committee has contacted a local network solutions provider with a request to sell for them a specified number of cables with equal lengths. The Judging Committee wants the cables to be as long as possible to sit contestants as far from each other as possible. 
  The Cable Master of the company was assigned to the task. He knows the length of each cable in the stock up to a centimeter,and he can cut them with a centimeter precision being told the length of the pieces he must cut. However, this time, the length is not known and the Cable Master is completely puzzled. 
  You are to help the Cable Master, by writing a program that will determine the maximal possible length of a cable piece that can be cut from the cables in the stock, to get the specified number of pieces.

  Input

  The first line of the input file contains two integer numb ers N and K, separated by a space. N (1 = N = 10000) is the number of cables in the stock, and K (1 = K = 10000) is the number of requested pieces. The first line is followed by N lines with one number per line, that specify the length of each cable in the stock in meters. All cables are at least 1 meter and at most 100 kilometers in length. All lengths in the input file are written with a centimeter precision, with exactly two digits after a decimal point.

  Output

  Write to the output file the maximal length (in meters) of the pieces that Cable Master may cut from the cables in the stock to get the requested number of pieces. The number must be written with a centimeter precision, with exactly two digits after a decimal point. 
  If it is not possible to cut the requested number of pieces each one being at least one centimeter long, then the output file must contain the single number "0.00" (without quotes).

  Sample Input

  4 11
  8.02
  7.43
  4.57
  5.39

  Sample Output

  2.00

• 分析：二分搜索。套用二分搜索的模型，令条件C(x):=可以得到K条长度为x的绳子，则问题变成了求满足C(x)条件的最大的x。在区间初始化时，只需使用充分大的数INF(>MAX_L)作为上界即可：lb=0, ub=INF。现在只要能高效地判断C(x)即可，由于长度为L_i的绳子最多可以切出floor(L_i/x)段长度为x绳子，因此C(x)=(floor(L_i/x)的总和是否大于或等于K)，这可以在O(n)内被判断出来
• 代码：

   #include <cstdio>
   #include <cctype>
   #include <algorithm>
   #include <cmath>
   #define num s-'0'
  
   using namespace std;
  
   const int MAX_N=;
   const int INF=;
   int n,k;
   double L[MAX_N];
  
   void read(int &x){
       char s;
       x=;
       bool flag=;
       while(!isdigit(s=getchar()))
           (s=='-')&&(flag=true);
       for(x=num;isdigit(s=getchar());x=x*+num);
       (flag)&&(x=-x);
   }
  
   void write(int x)
   {
       if(x<)
       {
           putchar('-');
           x=-x;
       }
       if(x>)
           write(x/);
       putchar(x%+'');
   }
  
   double search();
   bool C(double x);
  
   int main()
   {
       read(n);read(k);
       for (int i=; i<n; i++) scanf("%lf", &L[i]);
       double p = search();
       printf("%.2f", floor(p*)/);
       putchar('\n');
   }
  
   bool C(double x)
   {
       int sum=;
       for (int i=; i<n; i++)
       {
           sum+=(int)(L[i]/x);
           if (sum>=k) return true;
       }
       return false;
   }
  
   double search()
   {
       double lb=, ub=INF;
       //while (ub-lb>0.001)
       for (int i=; i<; i++)
       {
           double mid=(lb+ub)/;
           if (C(mid)) lb=mid;
           else ub=mid;
       }
       return ub;
   } 

  Cable master

• PS：关于二分搜索法的结束的判定，上面的代码指定了循环次数作为终止条件，1次循环可以把区间的范围缩小一半，100次循环则可以达到10^-30的精度范围，基本上是没有问题的，除此之外，也可以把终止条件设为像上面注释中(ub-lb)>EPS那样，指定一个区间的大小，在这种情况下，如果EPS取得太小了，可能会因为浮点小数精度的原因导致陷入死循环，要小心这一点。

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最大化最小值

Aggressive cows(POJ 2456)

• 原题如下：
  Aggressive cows

  Time Limit: 1000MS	Memory Limit: 65536K
  Total Submissions: 20518	Accepted: 9737

  Description

  Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,xN (0 <= xi <= 1,000,000,000).

  His C (2 <= C <= N) cows don't like this barn layout and become aggressive towards each other once put into a stall. To prevent the cows from hurting each other, FJ want to assign the cows to the stalls, such that the minimum distance between any two of them is as large as possible. What is the largest minimum distance?

  Input

  * Line 1: Two space-separated integers: N and C

  * Lines 2..N+1: Line i+1 contains an integer stall location, xi

  Output

  * Line 1: One integer: the largest minimum distance

  Sample Input

  5 3
  1
  2
  8
  4
  9

  Sample Output

  3

  Hint

  OUTPUT DETAILS:

  FJ can put his 3 cows in the stalls at positions 1, 4 and 8, resulting in a minimum distance of 3.

  Huge input data,scanf is recommended.

• 分析：类似的最大化最小值或者最小化最大值的问题，通常用二分搜索法解决。
  我们定义：C(d):=可以安排牛的位置使得最近的两头牛的距离不小于d，那么问题就变成了求满足C(d)的最大的d。另外，最近的间距不小于d也可以说成是所有牛的间距都不小于d，因此就有C(d)=可以安排牛的位置使得任意的牛的间距都不小于d，这个问题的判断使用贪心法就可以解决。
• 代码：

   #include <cstdio>
   #include <cctype>
   #include <algorithm>
   #include <cmath>
   #define num s-'0'
  
   using namespace std;
  
   const int MAX_N=;
   const int INF=0x3f3f3f3f;
   int N,M;
   int x[MAX_N];
  
   void read(int &x){
       char s;
       x=;
       bool flag=;
       while(!isdigit(s=getchar()))
           (s=='-')&&(flag=true);
       for(x=num;isdigit(s=getchar());x=x*+num);
       (flag)&&(x=-x);
   }
  
   void write(int x)
   {
       if(x<)
       {
           putchar('-');
           x=-x;
       }
       if(x>)
           write(x/);
       putchar(x%+'');
   }
  
   bool C(int);
  
   int main()
   {
       read(N);read(M);
       for (int i=; i<N; i++) read(x[i]);
       sort(x, x+N);
       int lb=, ub=INF;
       while (ub-lb>)
       {
           int mid=(lb+ub)/;
           if (C(mid)) lb=mid;
           else ub=mid;
       }
       write(lb);
       putchar('\n');
   }
  
   bool C(int d)
   {
       int last=;
       for (int i=; i<M; i++)
       {
           int crt=last+;
           while (crt<N && x[crt]-x[last]<d) crt++;
           if (crt==N) return false;
           last=crt;
       }
       return true;
   }

  Aggressive cows

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最大化平均值

• 问题描述：有n个物品的重量和价值分别是w_i和v_i。从中选出k个物品使得单位重量的价值最大。
• 限制条件：
  1≤k≤n≤10⁴
  1≤w_i，v_i≤10⁶
• 分析：定义：条件C(x):=可以选择使得单位重量的价值不小于x，因此，原问题就变成了求满足C(x)的最大的x。接下来就是C(x)可行性的判断了，假设选了某个物品的集合S，那么它们的单位重量的价值是∑v_i/∑w_i，因此就是判断是否存在S满足∑v_i/∑w_i≥x，将不等式变形，得到∑(v_i-x*w_i)≥0，因此，可以对(v_i-x*w_i)的值进行排序贪心地进行选取，故C(x)=((v_i-x*w_i)从大到小排列中的前k个的和不小于0)，每次判断的复杂度是O(nlogn)
• 代码：

   #include <cstdio>
   #include <cctype>
   #include <algorithm>
   #include <cmath>
   #define num s-'0'
  
   using namespace std;
  
   const int MAX_N=;
   const int INF=0x3f3f3f3f;
   int n,k;
   int v[MAX_N],w[MAX_N];
   double y[MAX_N];
  
   void read(int &x){
       char s;
       x=;
       bool flag=;
       while(!isdigit(s=getchar()))
           (s=='-')&&(flag=true);
       for(x=num;isdigit(s=getchar());x=x*+num);
       (flag)&&(x=-x);
   }
  
   void write(int x)
   {
       if(x<)
       {
           putchar('-');
           x=-x;
       }
       if(x>)
           write(x/);
       putchar(x%+'');
   }
  
   bool C(double);
  
   int main()
   {
       read(n);read(k);
       for (int i=; i<n; i++)
       {
           read(w[i]);
           read(v[i]);
       }
       double lb=, ub=INF;
       for (int i=; i<; i++)
       {
           double mid=(lb+ub)/;
           if (C(mid)) lb=mid;
           else ub=mid;
       }
       printf("%.2f\n",ub);
   }
  
   bool C(double x)
   {
       for (int i=; i<n; i++)
       {
           y[i]=v[i]-x*w[i];
       }
       sort(y,y+n);
       double sum=;
       for (int i=; i<k; i++)
       {
           sum+=y[n--i];
       }
       return sum>=;
   }

  最大化平均值